Mock Unit Test Analysis for Year 10 Cambridge Mathematics | 剑桥Year 10数学单元测试模拟卷解析

📚 Mock Unit Test Analysis for Year 10 Cambridge Mathematics | 剑桥Year 10数学单元测试模拟卷解析

This article presents a detailed walkthrough of a mock unit test designed for Year 10 students following the Cambridge IGCSE Mathematics syllabus. The test focuses on core Algebra and Functions topics, including solving equations, inequalities, composite and inverse functions, exponential equations, algebraic fractions, and graph transformations. Every question is accompanied by a step-by-step solution in both English and Chinese, highlighting common mistakes and effective strategies. Use this analysis to strengthen your understanding and boost your confidence before the real test.

本文为学习剑桥IGCSE数学课程的Year 10学生提供一份单元模拟测试的详细解析。测试内容聚焦代数与函数的核心主题，涵盖方程求解、不等式、复合函数与反函数、指数方程、代数分式以及图像变换。每道题目都配有中英双语的逐步解析，指出常见错误和高效解题策略。通过这份解析巩固理解，在真实考试前增强信心。

1. Solving a Linear Equation | 解一元一次方程

Question: Solve 2x − 5 = 3x + 7.

题目：解方程 2x − 5 = 3x + 7。

We want to collect the x terms on one side. Subtract 2x from both sides to get −5 = x + 7.

步骤1：将含 x 的项集中到一边。两边同时减去 2x，得到 −5 = x + 7。

Step 2: Subtract 7 from both sides to isolate x. This yields −5 − 7 = x, so x = −12.

步骤2：两边减去 7，使 x 单独出现。得到 −5 − 7 = x，因此 x = −12。

5 = −29; right side 3(−12) + 7 = −36 + 7 = −29. Both sides match, confirming the solution.

步骤3：将 x = −12 代回原方程进行验证：左边 2(−12) − 5 = −24 −5 = −29；右边 3(−12) + 7 = −36 + 7 = −29。两边相等，确认解正确。

Common mistake: When moving terms, some students forget to change the sign. Always perform the same operation on both sides and double-check.

常见错误：移项时，部分学生忘记变号。务必在等式两边执行相同运算，并仔细复查。

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2. Solving a Quadratic Equation by Factorising | 因式分解法解二次方程

Question: Solve x² − 4x − 12 = 0.

题目：解方程 x² − 4x − 12 = 0。

to −12 and add to −4. The pair −6 and 2 satisfies this: (−6) × 2 = −12 and (−6) + 2 = −4.

步骤1：对二次式进行因式分解。寻找两个数，乘积为 −12，和为 −4。−6 和 2 符合要求：(−6) × 2 = −12，(−6) + 2 = −4。

Step 2: Write the factorised form as (x − 6)(x + 2) = 0.

步骤2：写成因式分解形式 (x − 6)(x + 2) = 0。

to zero: x − 6 = 0 gives x = 6; x + 2 = 0 gives x = −2. The solutions are x = 6 or x = −2.

步骤3：令每个因式等于零：x − 6 = 0 得 x = 6；x + 2 = 0 得 x = −2。解为 x = 6 或 x = −2。

Product = −12	Sum = −4
−6 and 2	−4

表格总结了寻找因数的过程。

Common mistake: Students sometimes try to use the quadratic formula unnecessarily when simple factorisation is possible. Practise number pairs to become faster at factorising.

常见错误：学生有时在可以进行简单因式分解时仍使用求根公式。多练习数对搭配，可提高分解速度。

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3. Solving Linear Inequalities | 解一元一次不等式

Question: Solve the inequality 3x − 2 > 5x + 8.

题目：解不等式 3x − 2 > 5x + 8。

Step 1: Bring the x terms to one side. Subtract 3x from both sides: −2 > 2x + 8.

步骤1：将含 x 的项移到一边。两边同时减去 3x：−2 > 2x + 8。

Step 2: Subtract 8 from both sides: −10 > 2x.

步骤2：两边减去 8：−10 > 2x。

Step 3: Divide both sides by 2. Since we are dividing by a positive number, the inequality sign stays the same: −5 > x, which is equivalent to x < −5.

步骤3：两边除以 2。因为除以正数，不等号方向不变：−5 > x，等价于 x < −5。

Step 4: Express the solution as x < −5. Remember that dividing or multiplying by a negative number would reverse the sign, but here we divided by +2.

步骤4：将解表示为 x < −5。记住乘除负数才会改变不等号方向，这里是除以 +2，方向不变。

Common mistake: Forgetting to reverse the inequality when multiplying or dividing by a negative number. Always check the sign of the coefficient of x.

常见错误：乘除负数时忘记反转不等号。一定要检查 x 系数的正负。

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4. Composite Functions | 复合函数

Question: Given f(x) = 2x + 1 and g(x) = x² − 3, find fg(x) and gf(x), simplifying your answers.

题目：已知 f(x) = 2x + 1，g(x) = x² − 3，求 fg(x) 和 gf(x)，并化简结果。

tation fg(x) means f(g(x)). Substitute g(x) into f: f(g(x)) = 2(g(x)) + 1 = 2(x² − 3) + 1.

步骤1：记号 fg(x) 表示 f(g(x))。将 g(x) 代入 f：f(g(x)) = 2(g(x)) + 1 = 2(x² − 3) + 1。

Step 2: Simplify: 2x² − 6 + 1 = 2x² − 5. Thus fg(x) = 2x² − 5.

步骤2：化简：2x² − 6 + 1 = 2x² − 5。因此 fg(x) = 2x² − 5。

gf(x) which means g(f(x)). Substitute f(x) into g: g(f(x)) = (f(x))² − 3 = (2x + 1)² − 3.

步骤3：接着求 gf(x)，即 g(f(x))。将 f(x) 代入 g：g(f(x)) = (f(x))² − 3 = (2x + 1)² − 3。

Step 4: Expand (2x + 1)² = 4x² + 4x + 1. Then subtract 3: 4x² + 4x + 1 − 3 = 4x² + 4x − 2.

步骤4：展开 (2x + 1)² = 4x² + 4x + 1，然后减 3：4x² + 4x + 1 − 3 = 4x² + 4x − 2。

Step 5: So gf(x) = 4x² + 4x − 2. Notice that fg(x) and gf(x) are not the same in general.

步骤5：所以 gf(x) = 4x² + 4x − 2。注意 fg(x) 和 gf(x) 通常不相等。

Function	Expression
fg(x) = f(g(x))	2x² − 5
gf(x) = g(f(x))	4x² + 4x − 2

g fg(x) as f(x) × g(x). The correct interpretation is applying one function after another.

常见错误：误将 fg(x) 理解为 f(x) × g(x)。正确理解是先后代入两个函数。

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5. Finding the Inverse Function | 求反函数

Question: Find the inverse function f⁻¹(x) for f(x) = (3x − 1) / 2.

题目：求函数 f(x) = (3x − 1) / 2 的反函数 f⁻¹(x)。

Step 1: Write the function as y = (3x − 1) / 2.

步骤1：将函数写成 y = (3x − 1) / 2。

Step 2: Swap x and y to begin finding the inverse: x = (3y − 1) / 2.

步骤2：交换 x 和 y 的位置，开始求反函数：x = (3y − 1) / 2。

Step 3: Solve for y. Multiply both sides by 2: 2x = 3y − 1.

步骤3：解出 y。两边乘以 2：2x = 3y − 1。

Step 4: Add 1 to both sides: 2x + 1 = 3y.

步骤4：两边加 1：2x + 1 = 3y。

Step 5: Divide by 3: y = (2x + 1) / 3. Therefore f⁻¹(x) = (2x + 1) / 3.

步骤5：除以 3：y = (2x + 1) / 3。因此 f⁻¹(x) = (2x + 1) / 3。

)) = x: f((2x+1)/3) = (3×((2x+1)/3) −1)/2 = (2x+1−1)/2 = 2x/2 = x, confirming correctness.

步骤6：验证 f(f⁻¹(x)) = x：f((2x+1)/3) = (3×((2x+1)/3) −1)/2 = (2x+1−1)/2 = 2x/2 = x，确认无误。

Common mistake: Forgetting to swap x and y at the start, leading to an algebraic expression that is not the inverse.

常见错误：一开始忘记交换 x 和 y，导致得到的不是真正的反函数表达式。

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6. Solving Exponential Equations | 解指数方程

Question: Solve 2^2x = 8^{x − 1}.

题目：解方程 2^2x = 8^{x − 1}。

Step 1: Express both sides with the same base. Notice that 8 = 2³, so 8^{x − 1} = (2³)^{x − 1} = 2^(3(x − 1)) = 2^{3x − 3}.

步骤1：将两边表示为同底数。注意到 8 = 2³，所以 8^{x − 1} = (2³)^{x − 1} = 2^(3(x − 1)) = 2^{3x − 3}。

Step 2: The equation becomes 2^2x = 2^{3x − 3}. Since the bases are equal, we can set the exponents equal: 2x = 3x − 3.

步骤2：方程变为 2^2x = 2^{3x − 3}。由于底数相同，指数也相等：2x = 3x − 3。

Step 3: Solve the linear equation: subtract 2x from both sides to get 0 = x − 3, so x = 3.

步骤3：解这个一元一次方程：两边减去 2x，得到 0 = x − 3，故 x = 3。

Step 4: Check: left side 2^2×3 = 2⁶ = 64; right side 8³⁻¹ = 8² = 64. The solution is valid.

步骤4：检验：左边 2^2×3 = 2⁶ = 64；右边 8³⁻¹ = 8² = 64。解成立。

Common mistake: Attempting to take logarithms unnecessarily when the bases can be made equal. First try to rewrite to the same base.

常见错误：在底数可以化为相同时仍使用对数。应首先尝试写成同底数的形式。

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7. Simplifying Algebraic Fractions | 代数分式的化简

Question: Simplify fully (3x² − 12) / (x² − x − 6).

题目：完全化简 (3x² − 12) / (x² − x − 6)。

Step 1: Factorise the numerator. 3x² − 12 = 3(x² − 4) = 3(x − 2)(x + 2).

步骤1：对分子因式分解。3x² − 12 = 3(x² − 4) = 3(x − 2)(x + 2)。

Step 2: Factorise the denominator. x² − x − 6 = (x − 3)(x + 2).

步骤2：对分母因式分解。x² − x − 6 = (x − 3)(x + 2)。

Step 3: The fraction becomes [3(x − 2)(x + 2)] / [(x − 3)(x + 2)]. Cancel the common factor (x + 2), but note that x ≠ −2 to avoid division by zero.

步骤3：分式变为 [3(x − 2)(x + 2)] / [(x − 3)(x + 2)]。约去公因式 (x + 2)，但要注明 x ≠ −2，避免分母为零。

Step 4: The simplified expression is 3(x − 2) / (x − 3) or (3x − 6) / (x − 3), with the restriction x ≠ −2 and x ≠ 3.

步骤4：化简结果为 3(x − 2) / (x − 3) 即 (3x − 6) / (x − 3)，定义域要求 x ≠ −2 且 x ≠ 3。

Common mistake: Cancelling terms without factorising first, or forgetting to state the excluded values. Always factorise completely before cancelling.

常见错误：未先因式分解就直接约分，或忘记注明定义域限制。务必先完全分解再约分。

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8. Graph Transformations | 函数图像变换

Question: ed by the vector (2, −3). Given that f(x) = x², find the equation of the translated graph.

题目：函数 y = f(x) 的图像按向量 (2, −3) 平移。已知 f(x) = x²，求平移后图像的方程。

Step 1: A translation by vector (h

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